Question

If
$\rm \sin θ + \sin ϕ =a$
$\rm \cos θ + \cos ϕ =b$
the find the value of,
$\rm \tan \bigg( \dfrac{θ - ϕ}{2} \bigg)$
​

Answer 1

Solution :-

Given that

sin θ + sin ϕ = a --------(1)

On squaring both sides then

(sin θ + sin ϕ)² = a²

=> sin² θ+sin² ϕ+ 2 sin θ sin ϕ = a²------(2)

Since , (a+b)² = a²+2ab+b²

And

cos θ + cos ϕ = b -------(3)

On squaring both sides then

=> cos² θ+cos² ϕ+2 cos θ cos ϕ = b²---(4)

Since , (a+b)² = a²+2ab+b²

On adding (2)&(4)

sin² θ + sin² ϕ + 2 sin θ sin ϕ + cos² θ + cos² ϕ + 2 cos θ cos ϕ = a²+b²

=> (sin² θ + cos² θ) + (sin² ϕ + cos² ϕ) + 2 sin θ sin ϕ + 2 cos θ cos ϕ = a²+b²

=> 1+1+2sin θ sin ϕ+ 2 cos θ cos ϕ = a²+b²

=> 2+2sin θ sin ϕ+ 2 cos θ cos ϕ = a²+b²

=> 2+2(sin θ sin ϕ+ cos θ cos ϕ ) = a²+b²

=> 2+2[ cos (θ -ϕ)] = a²+b²

Since,cos(A-B)=cos A cos B+sin A sin B

=> 2[ 1 + {cos (θ -ϕ)} ] = a²+b²

=> 1 + {cos (θ -ϕ)} = (a²+b²)/2

=> 2 cos² (θ -ϕ)/2 = (a²+b²)/2

=> cos² (θ -ϕ)/2 = (a²+b²)/(2×2)

=> cos² (θ -ϕ)/2 = (a²+b²)/4 -------(5)

We know that

tan (θ -ϕ)/2 = √[{1-cos²(θ -ϕ)/2 }/(cos²(θ -ϕ)/2 )]

=> tan (θ -ϕ)/2

=> √[{1-(a²+b²)/4}/{(a²+b²)/4}]

=> √[{4-(a²+b²)/4}/{(a²+b²)/4}]

=> √[{(4-a²-b²)/4}/{(a²+b²)/4}]

=> √[{(4-a²-b²)/4}×{4/(a²+b²)}]

=> √(4-a²-b²)/(a²+b²)

Therefore,

tan (θ -ϕ)/2 = √(4-a²-b²)/(a²+b²)

Answer :-

The value of tan (θ -ϕ)/2 =

√(4-a²-b²)/(a²+b²)

Used formulae:-

→ (a+b)² = a²+2ab+b²

→ cos (A-B) = cos A cos B + sin A sin B

→ tan (A-B)/2

= √[{(1-cos²(A-B)/2}/{cos²(A-B)/2}]

Answer 2

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Given that

(1)

On squaring both sides then

sin 8-sin-a

(sin e + sin )² = 8²

→sin¹ etsin² + 2 sin 8 sin =a² (2) Since (a+b)=a+2ab+b²

And

cos 8 + cos = b (3)

On squaring both sides then

cos² 0+cos³ p+2 cos e cos=b²-(4)

Since, (a+b)² = a¹+2ab+b²

On adding (2)S(4)

sin² 0+s

+ sin² + 2 sin 8 sin + cos² 0+

cos² + 2 cos e cos=a²+b²

→ (sin e + cos² e) + (sin²+ cos² ) - 2

sin e sin + 2 cos e cos = a +b² →> 1+1+2sin 8 sin + 2 cos e cos=a²+b²

→> 2+2sin 8 sin + 2 cos 8 cos=a²+b²

→> 2+2(sin 8 sin + cos 8 cos ) =a²+b² → 2+2[ cos (8-0)]=a²+b³

Since.cos(A-B)-cos A cos B+sin A sin B 2[1+ (cos (9-))]=a²+b²

1+ (cos (0-0)) = (a+b)/2

>>2 cas² (8-0)/2 = (a²+b³/2

>>cos² (8-0)/2 = (a+b)/(2*2)

>cos² (8-b/2 = (a+b)/4 (5)

We know that

tan (e-p)/2 = √(1-cos³(0-0)/2 1/(cos(8

→> v[[4-(a+b)/43/[(a+b)/4]] => [(4-a²-b³)/4)/((a+b)/4]]

→ √(4-a³-b³/(a+b)

Therefore,

tan (0-0)/2-(4-a²-b³)/(a+b)

Answer :

The value of tan (9-b)/2 =

√(4-a²-b³)/(a+b) Used formulae:

(a+b)²=a²+2ab+b²

- cos (A-B) = cos A cos B+ sin A sin B

-tan (A-B)/2

= [[(1-cos (A-B)/2}/{cos (A-B)/2]]